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If $f(x)$ is a function such that $f^{\prime \prime}(x)+f(x)=0$ and $g(x)=[f(x)]^{2}+\left[f^{\prime}(x)\right]^{2}$ and $g(3)+8$, then $g(8)=$
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The correct answer is:
8
We have,
$g(x)=[f(x)]^{2}+\left[f^{\prime}(x)\right]^{2}$
Differentiate the function $g(x)$
$g^{\prime}(x)=2 f(x) f^{\prime}(x)+2 f^{\prime}(x) f^{\prime \prime}(x)$
Use chain rule,
$2 f^{\prime}(x)\left[f(x)+f^{\prime \prime}(x)\right]=2 f^{\prime}(x)(0)=0$
Hence, $g(x)$ is $a$ constant function $\Rightarrow \quad g(x)=c$, constant
But, $g(3)=8$, so $g(x)=8$
For all real $x$.
Hence, $g(8)=8$
$g(x)=[f(x)]^{2}+\left[f^{\prime}(x)\right]^{2}$
Differentiate the function $g(x)$
$g^{\prime}(x)=2 f(x) f^{\prime}(x)+2 f^{\prime}(x) f^{\prime \prime}(x)$
Use chain rule,
$2 f^{\prime}(x)\left[f(x)+f^{\prime \prime}(x)\right]=2 f^{\prime}(x)(0)=0$
Hence, $g(x)$ is $a$ constant function $\Rightarrow \quad g(x)=c$, constant
But, $g(3)=8$, so $g(x)=8$
For all real $x$.
Hence, $g(8)=8$
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